SOLUTION: using matrices solve each system of equations x+y+z=0, 2x+3y+2z=-1, x-y+z=2

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Question 670080: using matrices solve each system of equations x+y+z=0, 2x+3y+2z=-1, x-y+z=2
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

By elimination

x+y+z=0

x-y+z=2

   2x +2z = 2   0r x+%2B+z+=+1

2x+3y+2z=-1 

 x-y+z=2

    5x + 5z = 5

     x +  z = 1

          0 ≠ 6   NO solution for this System




 
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables




  

  

  system%281%2Ax%2B1%2Ay%2B1%2Az=0%2C2%2Ax%2B3%2Ay%2B2%2Az=-1%2C1%2Ax%2B-1%2Ay%2B1%2Az=2%29

  

  

  

  First let A=%28matrix%283%2C3%2C1%2C1%2C1%2C2%2C3%2C2%2C1%2C-1%2C1%29%29. This is the matrix formed by the coefficients of the given system of equations.

  

  

  Take note that the right hand values of the system are 0, -1, and 2 and they are highlighted here: 

  

  

  

  

  These values are important as they will be used to replace the columns of the matrix A.

  

  

  

  

  Now let's calculate the the determinant of the matrix A to get abs%28A%29=0. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.

  

  

  

  Notation note: abs%28A%29 denotes the determinant of the matrix A.

  

  

  

  Since the determinant of matrix A is zero, this means that we cannot use Cramer's Rule. Why? Remember that each solution 'x', 'y', and 'z' are found by dividing by the determinant of A. If that determinant is zero, then you'll be dividing by zero, which is undefined. So that means you have to use an alternate method to find the solution.